Question

我的错误是什么

<?php
$conn = mysqli_connect("localhost","root","");
if(!$conn)
{
    die('Server not Connected : '.mysqli_error());  
}   
if(!mysqli_select_db($conn,'myphp'))
{
    echo "Database not connected";
}
if(isset($_POST['submit']))
// if(isset($_POST["f_name"]) && isset($_POST["s_name"]) && isset($_POST["age"]) && isset($_POST["email_id"]) && isset($_POST["dob"]))
{
    $first_name  = $_POST["f_name"];
    $second_name = $_POST["s_name"];
    $agee        = $_POST["age"];
    $email_ids    = $_POST["email_id"];
    $dobs         = $_POST["dob"];

    $sql = "INSERT INTO crud (First Name, Last Name, Age, Email Id, DOB) VALUES ('".$_POST["f_name"]."','".$_POST["s_name"]."','".$_POST["age"]."','".$_POST["email_id"]."','".$_POST["dob"]."')";
    }
 if(!mysqli_query($conn,$sql))
 {
     echo "Not Insert sucessfully";
 }
 else 
 {
     echo "Successfully Import th Data to Database";
 }  
 ?>
<html>
<head>
<title>My CRUD Model</title>
</head>
<body>

<form method="POST" action="">
First Name : <input type="text" name="f_name"/><br>
Second Name  : <input type="text" name="s_name"/><br>
Age        : <input type="number" name="age"/><br>
Email Id   : <input type="email" name="email_id"/><br>
DOB        : <input type="date" name="dob"/><br>
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>

Answer 1

请在文件名下面用数据库替换一个：如果无法正常工作，请发布有关您的桌子屏幕截图的问题

INSERT INTO crud（名字，姓氏，年龄，email_id，dob）

$sql = " INSERT INTO crud (first_name, last_name, age, email_id, dob) VALUES ('".$_POST["f_name"]."','".$_POST["s_name"]."','".$_POST["age"]."','".$_POST["email_id"]."','".$_POST["dob"]."')";

Answer 2

用以下代码替换php代码并检查。

<?php
$conn = mysqli_connect("localhost","root","","myphp");
if(!$conn)
{
    die('Database not connected : '.mysqli_error());  
}   
if(isset($_POST['submit']))
// if(isset($_POST["f_name"]) && isset($_POST["s_name"]) && isset($_POST["age"]) && isset($_POST["email_id"]) && isset($_POST["dob"]))
{
    $first_name  = $_POST["f_name"];
    $second_name = $_POST["s_name"];
    $agee        = $_POST["age"];
    $email_ids    = $_POST["email_id"];
    $dobs         = $_POST["dob"];

    $sql = "INSERT INTO crud (first_name, last_name, age, email_id, dob) VALUES ('".$_POST["f_name"]."','".$_POST["s_name"]."','".$_POST["age"]."','".$_POST["email_id"]."','".$_POST["dob"]."')";
     if(!mysqli_query($conn,$sql))
     {
         echo "Not Insert sucessfully";
     }
     else 
     {
         echo "Successfully Import th Data to Database";
     } 
 } 
 ?>

PHP-CRUD（插入）错误

2 个答案: